Dynamic System Response
• Solution of Linear, Constant-Coefficient, Ordinary Differential Equations – Classical Operator Method – Laplace Transform Method • Laplace Transform Properties • 1st-Order Dynamic System Time and Frequency Response • 2nd-Order Dynamic System Time and Frequency Response
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 1 Laplace Transform Methods
• A basic mathematical model used in many areas of engineering is the linear ordinary differential equation with constant coefficients: d nq d n-1q dq a o + a o + +a o + a q = n dt n n-1 dt n-1 L 1 dt 0 o d mq d m-1q dq b i + b i + +b i + b q m dt m m-1 dt m-1 L 1 dt 0 i
• qo is the output (response) variable of the system • qi is the input (excitation) variable of the system • an and bm are the physical parameters of the system
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 2 • Straightforward analytical solutions are available no matter how high the order n of the equation. • Review of the classical operator method for solving linear differential equations with constant coefficients will be
useful. When the input qi(t) is specified, the right hand side of the equation becomes a known function of time, f(t). • The classical operator method of solution is a three-step procedure:
– Find the complimentary (homogeneous) solution qoc for the equation with f(t) = 0.
– Find the particular solution qop with f(t) present.
– Get the complete solution qo = qoc + qop and evaluate the constants of integration by applying known initial conditions.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 3 • Step 1
– To find qoc, rewrite the differential equation using the differential operator notation D = d/dt, treat the equation as if it were algebraic, and write the system characteristic equation as:
n n-1 a nD + a n-1D +L+a1D + a 0 = 0
– Treat this equation as an algebraic equation in the unknown D and
solve for the n roots (eigenvalues) s1, s2, ..., sn. Since root finding is a rapid computerized operation, we assume all the roots are available and now we state rules that allow one to immediately
write down qoc: – Real, unrepeated root s : 1 s1t qoc = c1e
– Real root s2 repeated m times:
s2t s2t 2 s2t m s2t qoc = c0e + c1te + c2t e +L+cmt e
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 4 – When the a’s are real numbers, then any complex roots that might appear always come in pairs a ± ib: at qoc = ce sinabt + ff – For repeated root pairs a ± ib, a ± ib, and so forth, we have: at at qoc = c0e sinabt + f0f + c1te sinabt + f1f+L
– The c’s and f’s are constants of integration whose numerical values cannot be found until the last step.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 5 • Step 2
– The particular solution qop takes into account the "forcing function" f(t) and methods for getting the particular solution depend on the form of f(t). – The method of undetermined coefficients provides a simple method of getting particular solutions for most f(t)'s of practical interest. – To check whether this approach will work, differentiate f(t) over and over. If repeated differentiation ultimately leads to zeros, or else to repetition of a finite number of different time functions, then the method will work. – The particular solution will then be a sum of terms made up of each different type of function found in f(t) and all its derivatives, each term multiplied by an unknown constant (undetermined coefficient).
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 6 – If f(t) or one of its derivatives contains a term identical to a term in
qoc, the corresponding terms should be multiplied by t. – This particular solution is then substituted into the differential equation making it an identity. Gather like terms on each side, equate their coefficients, and obtain a set of simultaneous algebraic equations that can be solved for all the undetermined coefficients.
• Step 3
– We now have qoc (with n unknown constants) and qop (with no unknown constants). The complete solution qo = qoc + qop. The initial conditions are then applied to find the n unknown constants.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 7 • Certain advanced analysis methods are most easily developed through the use of the Laplace Transform. • A transformation is a technique in which a function is transformed from dependence on one variable to dependence on another variable. Here we will transform relationships specified in the time domain into a new domain wherein the axioms of algebra can be applied rather than the axioms of differential or difference equations. • The transformations used are the Laplace transformation (differential equations) and the Z transformation (difference equations). • The Laplace transformation results in functions of the time variable t being transformed into functions of the frequency-related variable s.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 8 • The Z transformation is a direct outgrowth of the Laplace transformation and the use of a modulated train of impulses to represent a sampled function mathematically. • The Z transformation allows us to apply the frequency- domain analysis and design techniques of continuous control theory to discrete control systems. • One use of the Laplace Transform is as an alternative method for solving linear differential equations with constant coefficients. Although this method will not solve any equations that cannot be solved also by the classical operator method, it presents certain advantages: – Separate steps to find the complementary solution, particular solution, and constants of integration are not used. The complete solution, including initial conditions, is obtained at once.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 9 – There is never any question about which initial conditions are needed. In the classical operator method, the initial conditions are evaluated at t = 0+, a time just after the input is applied. For some kinds of systems and inputs, these initial conditions are not the same as those before the input is applied, so extra work is required to find them. The Laplace Transform method uses the conditions before the input is applied; these are generally physically known and are often zero, simplifying the work. – For inputs that cannot be described by a single formula for their entire course, but must be defined over segments of time, the classical operator method requires a piecewise solution with tedious matching of final conditions of one piece with initial conditions of the next. The Laplace Transform method handles such discontinuous inputs very neatly. – The Laplace Transform method allows the use of graphical techniques for predicting system performance without actually solving system differential equations.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 10 • All theorems and techniques of the Laplace Transform derive from the fundamental definition for the direct Laplace Transform F(s) of the time function f(t): ¥ L f(t) = Fasf = f(t)e-stdt t > 0 z0 s = a complex variable = s + iw • This integral cannot be evaluated for all f(t)'s, but when it can, it establishes a unique pair of functions, f(t) in the time domain and its companion F(s) in the s domain. Comprehensive tables of Laplace Transform pairs are available. Signals we can physically generate always have corresponding Laplace transforms. When we use the Laplace Transform to solve differential equations, we must transform entire equations, not just isolated f(t) functions, so several theorems are necessary for this.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 11 • Linearity Theorem:
L a1f1atf + a 2f2atf = L a1f1atf + L a2f2 atf = a1F1asf+ a 2F2 asf • Differentiation Theorem: df LL O = sFasf - fa0f NMdt QP 2 Ld f O 2 df L = s Fasf - sfa0f- a0f NMdt2 QP dt n n-1 Ld f O n n-1 n-2 df d f L = s Fasf - s fa0f - s a0f- - a0f NMdt n QP dt L dt n-1 – f(0), (df/dt)(0), etc., are initial values of f(t) and its derivatives evaluated numerically at a time instant before the driving input is applied.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 12 • Integration Theorem: a-1f Fasf f a0f L fatfdt = + z s s n a- kf a- nf Fasf f a0f L f atf = n + å n-k+1 s k=1 s a- nf n a-0f where f atf = zLz fatfadtf and f atf = fatf
– Again, the initial values of f(t) and its integrals are evaluated numerically at a time instant before the driving input is applied.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 13 • Delay Theorem: – The Laplace Transform provides a theorem useful for the dynamic system element called dead time (transport lag) and for dealing efficiently with discontinuous inputs.
u(t) = 1.0 Þ t > 0
= Þ < -as u(t) 0 t 0 L f(t - a)u(t - a) = e F(s) u(t - a) = 1.0 Þ t > a u(t - a) = 0 Þ t < a
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 14 • Final Value Theorem:
– If we know Q0(s), q0(¥) can be found quickly without doing the complete inverse transform by use of the final value theorem.
limt®¥ f(t) = lims®0 sF(s) – This is true if the system and input are such that the output approaches a constant value as t approaches ¥. • Initial Value Theorem: – This theorem is helpful for finding the value of f(t) just after the input has been applied, i.e., at t = 0+. In getting the F(s) needed to apply this theorem, our usual definition of initial conditions as those before the input is applied must be used.
limt®0 f(t) = lims®¥ sF(s)
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 15 • Impulse Function d(t) d(t) = 0 Þ t ¹ 0 +e p(t) Approximating function for d(t)dt = 1Þ e > 0 z-e the unit impulse function 1/b d(t) = limb®0 p(t) du 1 L d(t) = LL O = sU(s) = s = 1.0 NM dt QP s b t – The step function is the integral of the impulse function, or conversely, the impulse function is the derivative of the step function. – When we multiply the impulse function by some number, we increase the “strength of the impulse”, but “strength” now means area, not height as it does for “ordinary” functions.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 16 • An impulse that has an infinite magnitude and zero duration is mathematical fiction and does not occur in physical systems. • If, however, the magnitude of a pulse input to a system is very large and its duration is very short compared to the system time constants, then we can approximate the pulse input by an impulse function. • The impulse input supplies energy to the system in an infinitesimal time.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 17 • Inverse Laplace Transformation – A convenient method for obtaining the inverse Laplace transform is to use a table of Laplace transforms. In this case, the Laplace transform must be in a form immediately recognizable in such a table. – If a particular transform F(s) cannot be found in a table, then we may expand it into partial fractions and write F(s) in terms of simple functions of s for which inverse Laplace transforms are already known. – These methods for finding inverse Laplace transforms are based on the fact that the unique correspondence of a time function and its inverse Laplace transform holds for any continuous time function.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 18 Time Response & Frequency Response 1st-Order Dynamic System Example: RC Low-Pass Filter
i i in R out Dynamic System Investigation of the e e RC Low-Pass Filter in C out
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 19 Measurements, Calculations, Which Parameters to Identify? Model Manufacturer's Specifications What Tests to Perform? Parameter ID
Physical Physical Math System Model Model
Assumptions Physical Laws Equation Solution: Experimental and Analytical Analysis Engineering Judgement Model Inadequate: and Numerical Modify Solution
Actual Predicted Dynamic Compare Dynamic Behavior Behavior
Make Design Design Modify Model Adequate, Model Adequate, Complete or Decisions Performance Inadequate Performance Adequate Augment Dynamic System Investigation
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 20 Zero-Order Dynamic System Model
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 21 Validation of a Zero-Order Dynamic System Model
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 22 1st-Order Dynamic System Model
t = time constant K = steady-state gain
t = t
Slope at t = 0
t Kq - Kq is is t q = qe& o = & o t0= t t Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 23 • How would you determine if an experimentally- determined step response of a system could be represented by a first-order system step response?
t æö- t Straight-Line Plot: qo(t) =-Kqis ç÷1e èø éùqto ( ) t log10 êú1- vs. t q(t)- Kq - Kq ois =-e t ëûis Kqis t Slope = -0.4343/t qt( ) - 1e-=o t Kqis
éùqto ( ) tt log10êú1-=-log10 e=-0.4343 ëûKqis tt
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 24 – This approach gives a more accurate value of t since the best line through all the data points is used rather than just two points, as in the 63.2% method. Furthermore, if the data points fall nearly on a straight line, we are assured that the instrument is behaving as a first-order type. If the data deviate considerably from a straight line, we know the system is not truly first order and a t value obtained by the 63.2% method would be quite misleading. – An even stronger verification (or refutation) of first-order dynamic characteristics is available from frequency- response testing. If the system is truly first-order, the amplitude ratio follows the typical low- and high- frequency asymptotes (slope 0 and –20 dB/decade) and the phase angle approaches -90° asymptotically.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 25 – If these characteristics are present, the numerical value of t is found by determining w (rad/sec) at the
breakpoint and using t = 1/wbreak. Deviations from the above amplitude and/or phase characteristics indicate non-first-order behavior.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 26 • What is the relationship between the unit-step response and the unit-ramp response and between the unit-impulse response and the unit-step response? – For a linear time-invariant system, the response to the derivative of an input signal can be obtained by differentiating the response of the system to the original signal. – For a linear time-invariant system, the response to the integral of an input signal can be obtained by integrating the response of the system to the original signal and by determining the integration constants from the zero-output initial condition.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 27 • Unit-Step Input is the derivative of the Unit-Ramp Input. • Unit-Impulse Input is the derivative of the Unit- Step Input. • Once you know the unit-step response, take the derivative to get the unit-impulse response and integrate to get the unit-ramp response.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 28 System Frequency Response
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 29 Bode Plotting of 1st-Order Frequency Response
dB = 20 log10 (amplitude ratio) decade = 10 to 1 frequency change octave = 2 to 1 frequency change
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 30 i i Analog Electronics: in R out RC Low-Pass Filter Time Response & e e Frequency Response in C out
éùein éùRCs+-1Réùeout êú= êúêú ëûiin ëûCs1- ëûiout
eout 11 === when i0out ein RCs+1t+s1
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 31 Time Response to Unit Step Input
1
0.9
0.8
0.7
0.6 0.5 R = 15 KW
Amplitude 0.4 C = 0.01 mF 0.3
0.2
0.1
0 0 1 2 3 4 5 6 7 8 -4 Time (sec) x 10 Time Constant t = RC
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 32 • Time Constant t – Time it takes the step response to reach 63% of the steady-state value
• Rise Time Tr = 2.2 t – Time it takes the step response to go from 10% to 90% of the steady-state value
• Delay Time Td = 0.69 t – Time it takes the step response to reach 50% of the steady-state value
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 33 R = 15 KW Frequency Response C = 0.01 mF
0 0
-5 -20
-10 -40 Gain dB -15 -60 Phase (degrees)
-20 -80
-25 -100 102 103 104 105 102 103 104 105 Frequency (rad/sec) Frequency (rad/sec) Bandwidth = 1/t e KKÐ0Ko out (iw) ===Ð-tan -1 wt ei1wt+ 222-12 in (wt+) 1Ðtan1wt(wt+)
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 34 • Bandwidth – The bandwidth is the frequency where the amplitude ratio drops by a factor of 0.707 = -3dB of its gain at zero or low-frequency. – For a 1st -order system, the bandwidth is equal to 1/ t. – The larger (smaller) the bandwidth, the faster (slower) the step response. – Bandwidth is a direct measure of system susceptibility to noise, as well as an indicator of the system speed of response.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 35 MatLab / Simulink Diagram Frequency Response for 1061 Hz Sine Input
t = 1.5E-4 sec
1 output tau.s+1 output First-Order Sine Wave Plant
input t input Clock time
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 36 Amplitude Ratio = 0.707 = -3 dB Phase Angle = -45°
Response to Input 1061 Hz Sine Wave 1 Input 0.8
0.6
0.4
0.2
0 amplitude -0.2 Output -0.4
-0.6
-0.8
-1 0 0.5 1 1.5 2 2.5 3 3.5 4 time (sec) -3 x 10
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 37 Time Response & Frequency Response 2nd-Order Dynamic System Example: 2-Pole, Low-Pass, Active Filter
R 4 R 7
C 5 R 1 R 3 - R 6 -
C 2 + e in +
e out
Dynamic System Investigation of the Two-Pole, Low-Pass, Active Filter
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 38 Measurements, Calculations, Which Parameters to Identify? Model Manufacturer's Specifications What Tests to Perform? Parameter ID
Physical Physical Math System Model Model
Assumptions Physical Laws Equation Solution: Experimental and Analytical Analysis Engineering Judgement Model Inadequate: and Numerical Modify Solution
Actual Predicted Dynamic Compare Dynamic Behavior Behavior
Make Design Design Modify Model Adequate, Model Adequate, Complete or Decisions Performance Inadequate Performance Adequate Augment Dynamic System Investigation
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 39 Physical Model Ideal Transfer Function
æR 7 öæö1 ç÷ç÷ e RRRCC out (s) = è6øèø1325 ein 2 æö1111 ss+ç÷+++ èøR3C2R1C2R4C2RRCC3425
R 4 R 7
C 5 R 1 R 3 - R 6 -
C 2 + e in +
e out
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 40 a 2nd-Order Dynamic w= 0 undamped natural frequency n @ a System Model 2 a1 z=@ damping ratio 2 2aa20 dq00dq a22 +a1+=a0q0bq0i dtdt b0 K@ = steady-state gain 2 a 12dq00z dq 0 22++=q0iKq wwnndtdt
Step Response of a 2nd-Order System
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 41 2 Step Response 12dq00V dq +z+=q0iKq of a ww22dtdt nn 2nd-Order System Underdamped éù 1 -zwn t 2-12 qo=Kqisnêú1-esinw1-zt+sin1-z z<1 2 ( ) ëûêú1-z
=éù-+w-wnt z= Critically Damped qoKq1isnëû(1t)e 1
éù2 2 z+z-1 (-z+z-w1t) n êú1e- Over- êú21z-2 damped qo=Kqis êú z>1 2 2 êúz-z-1 (-z-z-w1t) n + e êú2 ëû21z- Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 42 Frequency Response of a 2nd-Order System
Qo K (s) = 2 Laplace Transfer Function Qi s2sz 2 ++1 wwnn Sinusoidal Transfer Function
Q K2z o (iw) =Ðtan-1 Q 2 2 æöww i éùæöw4zw22 - n êú1-+ ç÷ww ç÷ww2 èøn ëûêúèønn
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 43 Frequency Response of a 2nd-Order System
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 44 -40 dB per decade slope Frequency Response of a 2nd-Order System
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 45 Some Observations • When a physical system exhibits a natural oscillatory behavior, a 1st-order model (or even a cascade of several 1st-order models) cannot provide the desired response. The simplest model that does possess that possibility is the 2nd-order dynamic system model. • This system is very important in control design. – System specifications are often given assuming that the system is 2nd order. – For higher-order systems, we can often use dominant pole techniques to approximate the system with a 2nd- order transfer function.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 46 • Damping ratio z clearly controls oscillation; z < 1 is required for oscillatory behavior. • The undamped case (z = 0) is not physically realizable (total absence of energy loss effects) but gives us, mathematically, a sustained oscillation at
frequency wn. • Natural oscillations of damped systems are at the
damped natural frequency wd, and not at wn. 2 wdn=w1-z • In hardware design, an optimum value of z = 0.64 is often used to give maximum response speed without excessive oscillation.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 47 • Undamped natural frequency wn is the major factor in response speed. For a given z response speed is
directly proportional to wn. • Thus, when 2nd-order components are used in
feedback system design, large values of wn (small lags) are desirable since they allow the use of larger loop gain before stability limits are encountered. • For frequency response, a resonant peak occurs for
z < 0.707. The peak frequency is wp and the peak amplitude ratio depends only on z.
2 K wpn=w1-2z= peak amplitude ratio 2z1-z2
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 48 • Bandwidth – The bandwidth is the frequency where the amplitude ratio drops by a factor of 0.707 = -3dB of its gain at zero or low-frequency. – For a 1st -order system, the bandwidth is equal to 1/ t. – The larger (smaller) the bandwidth, the faster (slower) the step response. – Bandwidth is a direct measure of system susceptibility to noise, as well as an indicator of the system speed of response. – For a 2nd-order system:
224 BW=wn 1-z2+2-z+z44
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 49 – As z varies from 0 to 1, BW varies from 1.55wn to 0.64wn. For a value of z = 0.707, BW = wn. For most design considerations, we assume that the bandwidth of a nd 2 -order all pole system can be approximated by wn.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 50 Kw2 G(s) = n s22+2sVw+w nn Location of Poles s=-Vw±i1w-V2 Of 1,2nn Transfer Function si1,2d=-s±w
-st æös y(t) =1-eç÷coswddt+wsint èøwd 1.8 tr » rise time wn 4.6 ts » settling time Vwn -pV 2 General All-Pole 1-V nd Mp =e (0£z<1) overshoot 2 -Order æöz Step Response »ç÷1- (0£z£0.6) èø0.6 Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 51 1.8 4.6 w³n s³ t r ts
z³0.6(1-Mp ) 0£z£0.6
Time-Response Specifications vs. Pole-Location Specifications
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 52 Experimental Determination of z and wn
• z and wn can be obtained in a number of ways from step or frequency-response tests. • For an underdamped second-order system, the
values of z and wn may be found from the relations: pz - 1-z2 1 Mep =Þ z= 2 æöp ç÷+1 ç÷ èølogMep( )
2 wd 2p 2p wd=wnn1-zÞ w== T = 1-z22T1-z wd
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 53 • Logarithmic Decrement d is the natural logarithm of the ratio of two successive amplitudes. æöxt( ) d=ln=lneTzwn T =zw ç÷( ) n èøx(tT+ ) zw22pzwp 2pz =nn== w 22 d wn 11-z-z d z= Free Response of a 2nd-Order System 2p2 +d2 ( ) 2p T = 1 B1 w d= ln d nBn1+
-zwn t x(t) =Besint(wd +f)
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 54 – If several cycles of oscillation appear in the record, it is more accurate to determine the period T as the average of as many distinct cycles as are available rather than from a single cycle. – If a system is strictly linear and second-order, the value of n is immaterial; the same value of z will be found for any number of cycles. Thus if z is calculated for, say, n = 1, 2, 4, and 6 and different numerical values of z are obtained, we know that the system is not following the postulated mathematical model. • For overdamped systems (z > 1.0), no oscillations
exist, and the determination of z and wn becomes more difficult. Usually it is easier to express the system response in terms of two time constants.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 55 – For the overdamped step response:
éù2 2 z+z-1 (-z+z-w1t) n êú1e- êú21z-2 qo=Kqis êú z>1 2 2 êúz-z-1 (-z-z-w1t) n + e êú2 ëû21z- tt -- q tttt o =-+12e12e1 Kqist2-t1t21-t
11 – where tt12@@ z-z22-11wz+z-w ( ) nn( )
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 56 – To find t1 and t2 from a step-function response curve, we may proceed as follows:
• Define the percent incomplete response Rpi as:
æöqo Rpi @ ç÷1- 100 èøKqis
• Plot Rpi on a logarithmic scale versus time t on a linear scale. This curve will approach a straight line for large t if the system is
second-order. Extend this line back to t = 0, and note the value P1 where this line intersects the Rpi scale. Now, t1 is the time at which the straight-line asymptote has the value 0.368P1. • Now plot on the same graph a new curve which is the difference
between the straight-line asymptote and Rpi. If this new curve is not a straight line, the system is not second-order. If it is a straight line, the time at which this line has the value
0.368(P1-100) is numerically equal to t2.
• Frequency-response methods may also be used to find t1 and t2.
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 57 Step-Response Test for Overdamped Second-Order Systems
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 58 Frequency-Response Test of Second-Order Systems
Sensors & Actuators in Mechatronics K. Craig Dynamic System Response 59